-1.#J output

[maxzoran]

Anyone have a clue as to why this would be happening?

here's the code:

void solve_SSS (double elmnt_1, double elmnt_2, double elmnt_3){  /*begin SSS solver*/

	double ttl = 180;  /*initialize ttl as 180*/
	double A_num;  /*set angle A numerator as double*/
	double A_denom;  /*set angle A denominator as double*/
	double B_num;  /*set angle B numerator as double*/
	double B_denom;  /*set angle B denominator as double*/
	triangle s_SSS;  /*triangle typedef*/

	s_SSS.side_a = elmnt_1;  /*side a = elmnt_1*/
	s_SSS.side_b = elmnt_2,  /*side b = elmnt_2*/
	s_SSS.side_c = elmnt_3;  /*side c = elmnt_3*/
	A_num = (pow(s_SSS.side_b,2)+pow(s_SSS.side_c,2)-pow(s_SSS.side_a,2));  /*calc A numerator*/
	A_denom = (2*s_SSS.side_b*s_SSS.side_c);  /*calc A denom*/
	s_SSS.ang_A = acos(A_num/A_denom)*deg;  /*calc angle A*/
	B_num = (pow(s_SSS.side_c,2)+pow(s_SSS.side_a,2)-pow(s_SSS.side_b,2));  /*calc B numerator*/
	B_denom = (2*s_SSS.side_a*s_SSS.side_c);  /*calc B denom*/
	s_SSS.ang_B = acos(B_num/B_denom)*deg;  /*calc angle B*/
	s_SSS.ang_C = ttl - s_SSS.ang_A - s_SSS.ang_B;  /*calc angle C*/
	
	/*printf ("%.2f\n", A_num);
	printf ("%.2f\n", A_denom);*/


	printf ("Your solved triangle:"); /*print solved triangle*/
	printf ("\n");
	printf ("angle A = %.2f\n", s_SSS.ang_A);
	printf ("angle B = %.2f\n", s_SSS.ang_B);
	printf ("angle C = %.2f\n", s_SSS.ang_C);
	printf ("side a = %.2f\n", s_SSS.side_a);
	printf ("side b = %.2f\n", s_SSS.side_b);
	printf ("side c = %.2f\n", s_SSS.side_c);

	return;  /*end function*/

[KBriggs]

Why what is happening?

[maxzoran]

I get an output of -1.#J for angle A, angle B, and Angle C

[itCbitC]

Is variable "deg" initialized anywhere?

[whiteflags]

-1.#J is a sign that you are getting +/-inf or NaN as a result. The J is a result of rounding to get the .2 part of %.2f. Plain old %f might tell you which. If you're just leaving doubles uninitialized, its probably NaN though. Be aware that other math errors (such as dividing by zero) can cause this in floating-point types.

[KBriggs]

In cases of "degenerate triangles" some sides will have 0 length, and you will get division by 0. Put in some error checking on all your denominator terms to make sure they are nonzero before you divide.

[maxzoran]

ahhh yes...it was a divide by zero problem. Thanks for the help everyone!